LSAT 29, Logic Game 3, Question 19, LSATHacks

I hate this question. Not only is it hard, but it’s tough to explain. This is one of the hardest questions on the LSAT, so bear with me.

It’s very possible to do this question slowly, of course. You might have eventually figured out how to do it by drawing all the possibilities. But that’s not useful when you only have 8:45 to do the game.

Think of all of the variables that have to be locked down to determine the order:

1. FH
2. KL
3. G
4. S

To be useful, an answer must lock as many of these into place as possible. Each answer gives us two new conditions. 

To lock everything into a fixed position, each new condition in the answer has to take care of two things from that list.

So if a condition tells us H is before F, that’s not very useful. It only takes care of one thing from the list.

Telling us that H is before L is useful. It gives us this order: FHLK.

We took care of two things from the list at once.

A is not useful, for example. It tells us that K is before L. That only affects one thing from the list.

B is also not useful. If H is before K, that automatically means F is before H. So the first condition was wasted.

C takes care of FH and KL, but it doesn’t tell us where to put G and S. 

D does it. If G is before F, that means G is before this entire chain:

S must go first, because G can’t go first. 

If G is before F, we know F is before H. So far, we have:

S – G – F – H

The second condition tells us that L is before J. That also tells us that K is before L. 

Since K is after H, we get this order:

S – G – F – H – K – L – J

D is CORRECT.

E doesn’t do the trick. It’s not useful to tell us that H is before F, that only deals with one condition from the list.

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